A. Option D
B. Option C
C. Option B
D. Option A
Answer: C,D
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NEW QUESTION: 1
Which two methods mitigate congestion in a SAN network? (Choose two.)
A. Configure the port monitor to allow categorization of a specific device as slow.
B. Configure ER_RDY to allow splitting of each ISL between switches into separate virtual links.
C. Configure the port channel to enable individual buffer-to-buffer credits.
D. Configure the flow control for the FC to use R_RDY.
E. Use the port-monitor command to detect slow drain devices.
Answer: A,E
Explanation:
Reference:
cisco_mds9000_interfaces_config_guide_8x/congestion_avoidance_isolation.html
NEW QUESTION: 2
When components in a system are arranged in parallel, system reliability be:
A. The product of the reliability of individual components
B. As good as the weakest component
C. As good as the best component
D. Better than the best component
Answer: D
Explanation:
Regarding parallel and serial reliability: for components arranged in parallel, the reliability is better than the best component and for components arranged in serial, the reliability of the system is worse than the worst component. When components are arranged in parallel, only one of the units need to be successful for the whole system to be a success (i.e. the system fails only when every unit fails). There is redundancy built into the system. In a series arrangement, the failure of any one of the components leads to the failure of the whole system. Here is an example: Suppose two components with reliability 0.3 and 0.6 are arranged in parallel; then the reliability of the system is given by: R = 1 - [(1-R1)(1-R2)] = 1 - (0.7)(0.4) = 0.72, which is better than both 0.7 and 0.4.If two components with reliability 0.3 and 0.6 are arranged in series, then the reliability of the system is given by: R = R1 * R2 = 0.3 * 0.6 = 0.18, which is lower than both 0.3 and 0.6.Also, a parallel configuration (because of its redundancy) gives you a higher reliability (.72)than a similar serial configuration (.18). [Another way to look into this is: If, 1stS -> 1st being successful, 1stF -> 1st being unsuccessful, and 2ndS -> 2nd being successful, 2ndF -> 2nd being unsuccessful, then Probability of 1stS + 2ndS = 0.3 * 0.6 = 0.18 Probability of 1stS + 2ndF = 0.3 * 0.4 = 0.12 Probability of 1stF + 2ndS = 0.7 * 0.6 = 0.42 Probability of 1stF + 2ndF =
0.7 * 0.4 = 0.28 So, when they are in parallel, Probability of success = Probability of at least one being successful = 0.18 + 0.12 + 0.42 = 0.72 And, when they are in series, Probability of success = probability of both being successful = 0.18] So, the answer to this question is (a).
NEW QUESTION: 3
A. Option D
B. Option C
C. Option B
D. Option A
Answer: C,D